即使只使用数组有各种限制，也可以大大简化代码。不需要检查 sets . 只是：
分配一个数组来存储联合的所有元素（即。， int[] tmp_union )，最坏情况下将是两个数组中的所有元素 array1 以及 array2 .
迭代 array1 把它们和 tmp_union 数组，将它们添加到 tmp_union 仅当它们尚未添加到该数组时。
重复2）对于 array2 .
在此过程中，请跟踪添加到 tmp_union 迄今为止的阵列（即。， added_so_far ). 最后，从 tmp_union 将数组放入新数组（即。， unionArray )只为并集元素分配了空间。代码如下所示：

public static int[] union(int[] array1, int[] array2){
    int[] tmp_union = new int[array1.length + array2.length];
    int added_so_far = add_unique(array1, tmp_union, 0);
        added_so_far = add_unique(array2, tmp_union, added_so_far);
    return copyArray(tmp_union, added_so_far);
}

private static int[] copyArray(int[] ori, int size) {
    int[] dest = new int[size];
    for(int i = 0; i < size; i++)
        dest[i] = ori[i];
    return dest;
}

private static int add_unique(int[] array, int[] union, int added_so_far) {
    for (int element : array)
        if (!is_present(union, added_so_far, element))
            union[added_so_far++] = element;
    return added_so_far;
}

private static boolean is_present(int[] union, int added_so_far, int element) {
    for (int z = 0; z < added_so_far; z++)
         if (element == union[z])
             return true;
    return false;
}

使用java的流可以使这变得非常简单：

public int[] union(int[] array1, int[] array2) {
    return Stream.of(array1, array2).flatMapToInt(Arrays::stream).distinct().toArray();
}

这样做会容易得多 Collection api或 Stream 应用程序编程接口。但是，您已经提到，您希望只使用数组而不导入任何类，这将需要几个冗长（尽管很简单）的处理单元。驱动逻辑的最重要的理论是如何（如下所示）计算并集：

n(A U B) = n(A) + n(B) - n(A ∩ B)

和

n(Only A) = n(A) - n(A ∩ B)
n(Only B) = n(B) - n(A ∩ B)

此解决方案的高级摘要如下图所示：

代码本身的注解已经非常清楚地提到了其余的逻辑。

public class Main {
    public static void main(String[] args) {
        // Test
        display(union(new int[] { 1, 2, 3, 4 }, new int[] { 3, 4, 5, 6 }));
        display(union(new int[] { 1, 2, 3 }, new int[] { 4, 5, 6 }));
        display(union(new int[] { 1, 2, 3, 4 }, new int[] { 1, 2, 3, 4 }));
        display(union(new int[] { 1, 2, 3, 4 }, new int[] { 3, 4 }));
        display(union(new int[] { 1, 2, 3, 4 }, new int[] { 4, 5 }));
        display(union(new int[] { 1, 2, 3, 4, 5, 6 }, new int[] { 7, 8 }));
    }

    public static int[] union(int[] array1, int[] array2) {
        // Create an array of the length equal to that of the smaller of the two array
        // parameters
        int[] intersection = new int[array1.length <= array2.length ? array1.length : array2.length];
        int count = 0;

        // Put the duplicate elements into intersection[]
        for (int i = 0; i < array1.length; i++) {
            for (int j = 0; j < array2.length; j++) {
                if (array1[i] == array2[j]) {
                    intersection[count++] = array1[i];
                }
            }
        }

        // Create int []union of the length as per the n(A U B) = n(A) + n(B) - n(A ∩ B)
        int[] union = new int[array1.length + array2.length - count];

        // Copy array1[] minus intersection[] into union[]
        int lastIndex = copySourceOnly(array1, intersection, union, count, 0);

        // Copy array2[] minus intersection[] into union[]
        lastIndex = copySourceOnly(array2, intersection, union, count, lastIndex);

        // Copy intersection[] into union[]
        for (int i = 0; i < count; i++) {
            union[lastIndex + i] = intersection[i];
        }

        return union;
    }

    static int copySourceOnly(int[] source, int[] exclude, int[] target, int count, int startWith) {
        int j, lastIndex = startWith;
        for (int i = 0; i < source.length; i++) {
            // Check if source[i] is present in intersection[]
            for (j = 0; j < count; j++) {
                if (source[i] == exclude[j]) {
                    break;
                }
            }

            // If j has reached count, it means `break;` was not executed i.e. source[i] is
            // not present in intersection[]
            if (j == count) {
                target[lastIndex++] = source[i];

            }
        }
        return lastIndex;
    }

    static void display(int arr[]) {
        System.out.print("[");
        for (int i = 0; i < arr.length; i++) {
            System.out.print(i < arr.length - 1 ? arr[i] + ", " : arr[i]);
        }
        System.out.println("]");
    }
}

输出：

[1, 2, 5, 6, 3, 4]
[1, 2, 3, 4, 5, 6]
[1, 2, 3, 4]
[1, 2, 3, 4]
[1, 2, 3, 5, 4]
[1, 2, 3, 4, 5, 6, 7, 8]